Q61 / 90 · JEE Main 2021 (20 Jul Shift 2) · Physics

If the kinetic energy of a moving body becomes four times its initial kinetic energy, then the percentage change in its momentum will be:

Solution

$K_{2} = 4 K_{1}$

$\frac{1}{2} m v_{2}^{2} = 4 \times \frac{1}{2} m v_{1}^{2}$

$v_{2} = 2 v_{1}$

$P = m v$

$P_{2} = m v_{2} = 2 m v_{1}$

$P_{1} = m v_{1}$

$%$ change $= \frac{Δ P}{P_{1}} \times 100 = \frac{2 m v_{1} - m v_{1}}{m v_{1}} \times 100 = 100 %$

Hence, the correct answer is option: 'A'.